
The time period of a freely suspended magnet is $2\,sec$. If it is broken in length into two equal parts and one part is suspended in the same way, then its time period will be:
A. $4\,sec$
B. $\sqrt 2 \,sec$
C. $2\,sec$
D. $1\,sec$
Answer
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Hint: We have an equation of time period of a magnet that connects the quantity of cuts and the magnet's original time period to get the new time period. Substitute the values in the equation to get the new time period for the magnet that is suspended.
Formula used:
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.
Complete step by step solution:
The suspended magnet’s initial time period is $2\,sec$ given in the question. In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement. As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration.
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
When we broke the suspended magnet in length into two equal parts, then we have
$T' = \dfrac{T}{n}$
Here, $n$ is the number of cuts.
From the given information, we have $n = 2$, so substitute the value of $n$ in the above equation, then we obtain:
$T' = \dfrac{2}{2} \\$
$\therefore T' = 1\,sec $
Therefore, the time period will be $1\,sec$.
Thus, the correct option is D.
Note:The ratio of the magnetomotive force required to produce flux density per unit length of a given material is known as the magnetic field strength or magnetic field intensity. The magnet's moment of inertia has a direct relationship with the oscillation period and deep inside the earth's core, the magnetic field of the planet is created. The creation of magnetic fields by the flow of liquid iron causes the Earth's core to produce an electric current.
Formula used:
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.
Complete step by step solution:
The suspended magnet’s initial time period is $2\,sec$ given in the question. In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement. As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration.
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
When we broke the suspended magnet in length into two equal parts, then we have
$T' = \dfrac{T}{n}$
Here, $n$ is the number of cuts.
From the given information, we have $n = 2$, so substitute the value of $n$ in the above equation, then we obtain:
$T' = \dfrac{2}{2} \\$
$\therefore T' = 1\,sec $
Therefore, the time period will be $1\,sec$.
Thus, the correct option is D.
Note:The ratio of the magnetomotive force required to produce flux density per unit length of a given material is known as the magnetic field strength or magnetic field intensity. The magnet's moment of inertia has a direct relationship with the oscillation period and deep inside the earth's core, the magnetic field of the planet is created. The creation of magnetic fields by the flow of liquid iron causes the Earth's core to produce an electric current.
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